Simple harmonic motion

Oscillation in which the **acceleration is proportional to the displacement** from equilibrium and is always directed **back toward equilibrium**.

$a = -\omega^{2}x$ — **given** in the data booklet. a = acceleration, ω = angular frequency, x = displacement.

The acceleration points **opposite to the displacement** — always back toward equilibrium (the restoring direction).

A force that always acts to push or pull the object **back toward its equilibrium (resting) position**.

The central resting position where the object would sit still — where the displacement x = 0.

1. Acceleration **proportional to** displacement. 2. Acceleration directed **back toward equilibrium** (opposite to x).

A **straight line through the origin** with a **negative slope** equal to −ω².

There is a **restoring force** (and acceleration) directed **back to equilibrium** that is **proportional to the displacement** — exactly the condition a = −ω²x.

The steady loss of energy from an oscillation (to friction or drag), so each successive swing has a **smaller amplitude**.

The **amplitude slowly decreases** over many cycles while the **period stays almost the same**.

Yes — same form as a = −ω²x, so ω² = 25 → **ω = 5.0 rad s⁻¹**.

F = −kx, so **a = −(k/m)x**: acceleration proportional to displacement, towards equilibrium (ω² = k/m).

A **straight line through the origin with a negative gradient**.

The extra upthrust is **proportional to how far it is pushed down** and points back up to equilibrium.

The **time for one complete oscillation** (one full cycle), measured in seconds.

The **number of oscillations per second**, measured in hertz (Hz). f = 1 ÷ T.

$f = \dfrac{1}{T}$ and $T = \dfrac{1}{f}$ — they are reciprocals. **Given** in the data booklet (T = 1/f).

$T = 2\pi\sqrt{\dfrac{m}{k}}$ — depends on the mass m and spring constant k. **Given**.

$T = 2\pi\sqrt{\dfrac{l}{g}}$ — depends on the length l and gravity g. **Given**.

**No** — mass does not appear in T = 2π√(l/g), so the period is unchanged.

**No** — g does not appear in T = 2π√(m/k); only the mass and stiffness matter.

**×√4 = ×2** — the period doubles, because T ∝ √l.

**×1/√2 ≈ 0.71** — a stiffer spring oscillates faster, so a shorter period (T ∝ 1/√k).

How fast the cycle turns (2π radians per cycle): **ω = 2πf = 2π ÷ T**. Unit: rad s⁻¹.

Both period formulas have a √, so a quantity ×4 inside the root comes out as ×√4 = ×2.

The spring's **stiffness** — a bigger k means a stiffer spring that pulls back harder and oscillates faster.

The **largest displacement** from the equilibrium position (in m).

After **a quarter of a period, T/4**, as it passes through the equilibrium position.

The stiffness doubles (2k), so **T falls by √2**: T₂ = T₁ ÷ √2.

All three are **sinusoids** (smooth waves), but **shifted** in phase relative to one another.

Velocity **leads** displacement by a **quarter-cycle (90°)** — v is biggest at the centre, zero at the ends.

They are **antiphase (180° apart)** — a is the mirror image of x. This is the rule **a = -ω²x**.

At the **centre** (equilibrium, x = 0). It is **zero** at the turning points (maximum displacement).

At the **turning points** (maximum displacement). It is **zero** at the centre, because a = -ω²x.

The acceleration always points **back toward the equilibrium position** (a restoring acceleration), opposite to the displacement.

**T/4** — one quarter of the period (each quarter-cycle takes the same time).

**T/2** — half a period (two quarter-cycles, passing through the centre).

$a = -\omega^{2}x$ — acceleration proportional to displacement and directed back toward equilibrium.

$T = \dfrac{1}{f} = \dfrac{2\pi}{\omega}$ — both given in the data booklet.

Thinking it is greatest at the ends — it is greatest at the **centre** and **zero** at the ends.

**Four** quarter-periods, each lasting **T/4**.

Where the graph is **flat**: at the peaks and troughs (the turning points).

**v = ±ω√(x₀² − x²)** (given). At x = 0 it is v max = ωx₀.

Where the graph is **steepest**: as it crosses x = 0 (equilibrium).

It sets **where the cycle starts**: φ = 0 from equilibrium, φ = π/2 from +x₀. In radians.

They **peak at the same time**: no time gap between them.

**Kinetic energy** (of motion) and **potential energy** (stored, e.g. in a stretched spring). They swap back and forth as it oscillates.

It **stays constant** — KE and PE just trade places, but their sum never changes.

At the **centre** (equilibrium position), where the object moves **fastest**.

At the **ends** (the amplitude), where the object is **momentarily at rest**.

The **greatest displacement** from the centre — the turning point where the object briefly stops.

$E_{total} = \tfrac{1}{2}kA^{2}$ — set by the amplitude A. It is the data booklet's elastic PE ½kΔx² at Δx = A.

At the amplitude the object is at rest (KE = 0), so all the energy is the elastic PE stored at the biggest stretch, ½kx² with x = A.

Set the **maximum KE equal to the total energy**: ½mv_{max}² = ½kA², then solve for v_{max}.

It becomes **four times larger**, because E_{total} = ½kA² depends on A² (the amplitude squared).

It **equals the total energy** — at the centre the PE is zero, so all the energy is kinetic.

**PE** is an upward parabola (min at the centre); **KE** is a downward parabola (max at the centre); their sum is a flat line.

Thinking the speed is greatest at the ends — it is greatest at the **centre**; at the ends the object is momentarily still.

**Twice every period**: each time it passes the centre, in either direction.

**KE is maximum and PE is zero**: all the energy is kinetic at the centre.

**Kinetic, elastic potential and gravitational potential** energy; their total stays constant.

$E_T = \tfrac{1}{2}m\omega^2 x_0^2$ — a **constant**, set by the amplitude. Unit: **J**.

$E_K = \tfrac{1}{2}m\omega^2(x_0^2 - x^2)$ — biggest at the centre. Unit: **J**.

$E_P = E_T - E_K = \tfrac{1}{2}m\omega^2 x^2$ — biggest at the extremes. Unit: **J**.

$E_T \propto x_0^2$ — **double the amplitude → four times** the energy.

$E_T \propto \omega^2$ — **double ω → four times** the energy.

At the **centre** (x = 0), where it equals the total energy E_T.

At the **extremes** (x = ±x₀), where it equals the total energy E_T.

The **total energy** E_T. Kinetic and potential just **interchange**: E_K + E_P = E_T always.

Subtract: $E_P = E_T - E_K$ — don't recompute from scratch.

Both formulas use **x²**, so x = +0.06 m and x = −0.06 m give the **same** energies.

$E_T = \tfrac{1}{2}(0.20)(25)(0.010) = 0.025$ J.

Use $E_K = \tfrac{1}{2}mv^2$ and solve for $v = \sqrt{2E_K/m}$.

**E_T = ½mω²x₀²** and **E_P = ½mω²x²**. The kinetic energy is E_T − E_P.

**One quarter**: E_P ∝ x², and (½)² = ¼. The other ¾ is kinetic.

**√(3/4) v max ≈ 0.87 v max**, because E_K = ¾E_T and E_K ∝ v².